I self learn differential geometry for some time. To deal with curved spaces I assume there is a mapping that maps local coordinates to ambient coordinates in an Euclidean space.For example the longitude and latitude on a sphere to actual 3D point on the surface of a sphere in 3D Euclidean space using Cartesian coordinate system. That helps me visualize the (2 dimensional) manifolds and the tangent vectors and the parallel transports of them, and makes it easier to see how does the metric (via dot product) and the covariant derivative (via the generalization of the directional derivative) works.

But concepts in differential geometry often defined in terms of the connection or other related terms I don't understand yet.

Is mapping to the euclidean space yield correct results in general, or is it just accidental that it works for eg. a sphere?

Did I already use a connection without not knowing it?

How can I get a grasp of the concept of the connection from the knowledge I already have?


Mike asked "How does a mapping from your space to Euclidean space get you parallel transport?"

My thinking is like this. It will be detailed a bit.

Let's assume I'm traveling on a manifold, bringing a telescopic pointing stick me. I register both my position $p^j(t)$ and the direction of the stick $s^j(t)$ as a function of time. The position is the ordinary latitude, longitude pair, while stick is a local direction vector eg. "south-east 1 meters long."

But Earth is a 3 dimensional object so assume there exist a function that maps latitude-longitude pair to a 3D coordinate let's call it $M_i(\vec p)$.

If you move so only one coordinate changes, you either move north or east, so partial derivatives of this mapping function $\partial_j M_i$ give you the local basis vectors for each point.

(I will use $D$ for regular and $\partial_i$ to denote partial derivatives. They will operate on the symbol right next to them, juxtaposition doesn't imply parenthesis.)

Since direction of your stick is expressed in this basis the actual vector is: $s^j(t)\partial_j M_i(\vec p(t))$.

Taking the time derivative of this you have:

$$ Ds^j(t)\partial_j M_i(\vec p(t)) + s^j(t)\partial_k\partial_j M_i(\vec p(t)) D p^k(t) $$

There will be no more differentiation so we can omit parameters and clutter and take the values at a point directly:

$$ Ds^j \partial_j M_i + s^j \partial_k\partial_j M_i Dp^k $$

This vector is still a 3D vector that can potentially point out of the tangent plane, so we need to map it to the tangent space because we cannot "feel" the extradimensional component of it.

A general vector in ambient space can be expressed like this:

$$ u_i = v^j \partial_j M_i + w_i $$

Where $v^j$ represents the tangent plane component in local coordinates, so use the basis to make it an ambient vector. $w_i$ represents the component that's perpendicular to the tangent plane (in ambient ones).

Dotting both sides with a basis vector, we can get rid of the perpendicular component:

$$ u_i \partial_k M_i = v^j \partial_j M_i \partial_k M_i $$

The $\partial_j M_i \partial_k M_i$ is actually the metric. So all we need is to get rid of it by multiplying both sides with its inverse to get the tangent vector:

$$ \begin{array}{rcl} u_i \partial_k M_i & = & v^j g_{jk} \\ u_i g^{kl} \partial_k M_i & = & v^j g_{jk} g^{kl} \\ u_i g^{kl} \partial_k M_i & = & v^l \\ \end{array} $$

Doing the same with the derivative I computed before:

$$ Ds^j \partial_j M_i \partial_l M_i g^{lm} + s^j \partial_k\partial_j M_i \partial_l M_i g^{lm} Dp^k $$

In the left term that clutter evaluates to the Kronecker-delta so disappears, the clutter on the right is the Christoffel symbol of the second kind, so finally:

$$ Ds^m + \Gamma_{jk}^m s^j Dp^k $$

This is rate of change of the stick with respect to time mapped to the current tangent space.

To make parallel transport I'll need to choose $\vec s(t)$ so the previous expression evaluates to $0$.

Close enough? Where is the connection here?

  • 1
    $\begingroup$ I don't quite understand. How does a mapping from your space to Euclidean space get you parallel transport? $\endgroup$
    – user98602
    Commented Mar 2, 2015 at 18:53
  • $\begingroup$ For the case of the sphere, you are right that there is very little need for the full-fledged tools of differential geometry. But this is because the sphere can be entirely covered (excluding one point) by a single chart. But there may be manifolds where charts are "very local" so to speak, so that mapping to a Euclidean space is only "locally useful". $\endgroup$ Commented Mar 2, 2015 at 20:46
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    $\begingroup$ @ChristopherA.Wong: It seems he's talking about embeddings into Euclidean space, not charts. A good answer here would talk about intrinsic vs extrinsic geometry, induced metric/connection, Nash embedding theorem, etc. $\endgroup$ Commented Mar 3, 2015 at 0:28
  • $\begingroup$ Ah, sorry. Well, in that case, it does seem possible to do as he proposes, only that construction of explicit embeddings does not seem trivial... $\endgroup$ Commented Mar 3, 2015 at 19:03
  • $\begingroup$ @AnthonyCarapetis I think a good answer would first tell me what connection is. Because I'm not sure, texts often paraphrase it. But I'm yet to see one that directly say like "here it is, this is a connection". After that I'd look for comparison between the embedding method and doing the same with connections, (if there is a difference). $\endgroup$
    – Calmarius
    Commented Mar 4, 2015 at 8:15


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